# Feynman's Trick for Approximating $e^x$

Feynman knew how to approximate $e^x$ for small values of $x$ by noting the fact that \begin{align*} & \log10=2.30\qquad\qquad\therefore e^{2.3}\approx10\\ & \log 2 = 0.693\qquad\qquad\therefore e^{0.7}\approx2. \end{align*}

And he could approximate small values by performing some mental math to get an accurate approximation to three decimal places. For example, approximating $e^{3.3}$, we have$$e^{3.3}=e^{2.3+1}\approx 10e\approx 27.18281\ldots$$But what I am confused is how Feynman knew how to correct for the small errors in his approximation. The actual value is$$e^{3.3}=27.1126\ldots$$Perhaps someone can explain it to me?

• I'm not sure what you mean by "Feynman knew how to correct for the small errors in his approximation." Can you give more details / a reference? – Qiaochu Yuan Jul 7 '17 at 2:06
• @Qiaochu Yuan it is in the famous book Surely, You're joking, Mr.Feynman. – dezdichado Jul 7 '17 at 2:20
• @QiaochuYuan So we approximated $e^{3.3}$ to be about $27.18281$. However, the actual value is $e^{3.3}=27.11263$ And somehow, Feynman knew to subtract that tiny bit off from $27.18$. I'm wondering how he did that. It's also in his autobiography – Crescendo Jul 7 '17 at 2:21
• @dezdichado Interestingly enough, if you look at all the one-star ratings on Amazon for his book, they're all about how "he's so full of himself and how he's self-centred". – Crescendo Jul 7 '17 at 2:23
• @crescendo he sure was a bit of full of himself. But that's what made his autobiography so interesting and eccentric. – dezdichado Jul 7 '17 at 2:34

According to the book (p. 124 here), he knew $\log 10 \approx 2.3026$. Feynman ostensibly used the trivial linear approximation of $e^x$:

$$e^x\approx 1+x$$

which works well for small values of $x$.

Thus:

$$e^{3.3} = e^1\cdot e^{2.3026-0.0026} \approx e\cdot10\cdot e^{-0.0026}\approx 10e(1-0.0026) = 27.1121\dots$$

The correction adds two more correct decimal places and is quite easy to compute by hand.

Feynman used a similar trick to compute cubed roots faster than a man with an abacus (p. 127).

• Hm... I see. However, I tried to approximate $e^3$ and got the value off.\begin{align*}e^{3} & =e^{2.3026+0.7-0.0026}\\ & =e^{2.3026}e^{0.7}e^{-0.0026}\\ & =10\cdot 2(e^{-0.0026}) \\ & \approx20(1-0.0026) \\ & =19.948\end{align*}However, evaluating with a calculator yields$$e^3=20.0855$$Am I making a fool's error? – Crescendo Jul 7 '17 at 17:41
• @Crescendo The difference $0.7-\log 2=0.0069\dots$ is in fact more significant that the difference $\log 10-2.3$ which causes the discrepancy you observe. Try: $$e^3=e^{2.3026}e^{-0.0026}e^{0.6931}e^{0.0069}\approx 10\cdot(1-0.0026)\cdot 2\cdot (1+0.0069) = 20.0856$$ – Argon Jul 7 '17 at 20:09
• So really, there wasn't really anything wrong with what I had above? – Crescendo Jul 7 '17 at 23:15
• @Crescendo Nothing wrong exactly, just the approximation of $\log 2$ was (relatively) bad. This is what caused the error. Notice that $$e^{2.3} = 9.9741\dots\approx 10(1-0.0026)=9.974$$ – Argon Jul 8 '17 at 18:52

As an enthusiast of mental math calculations, here is what I think he did. Since he knew the basic approximations like you mentioned, he surely must have memorized the magnitude of the error: $$10-e^{2.3} = 0.02517...$$

But this is approximately $\dfrac{1}{4}\cdot 10^{-1}$. So basically it means you divide by $4$ and shift the decimal by one.

Thus, he can see that a better approximation for $e^{3.3}$ would be $10e-\dfrac{e}{4}\dfrac{1}{10} = 27.1828... - 0.068.. = 27.1125.$

This method of mine takes literally couple of seconds and obtains an approximation almost the same as Feynman did, so perhaps he did something similar.

There is no magic in mental calculations - you just get used to seeing lots of numbers and figuring out how to deal with them. Like, it is better to divide by $4$ instead of multiplying by $0.25$, which is what I did above.

• Interesting... I certainly have never thought of it that way! – Crescendo Jul 7 '17 at 17:47